Vrftsjtryk

How to prove that these two sets are convex for certain $p$?
And for what p they will not be convex?
$$A= {(x,y) in Bbb R^2 : |x|^p+|y|^p le 1, p in Bbb R}$$
$$B = {(x,y) in Bbb R^2 : x>0,y>0, x^p+y^p le 1, p in Bbb R}$$

$forall x,y in A rightarrow {z in A: z=(1-lambda)x+lambda y in A, lambda in [0,1] }$

I do not understand how the definition will look like applied to the condition of the set

Obviously your problem is understanding how the set is defined:

The set does not contain elements $x$ and $y$, but it cointas elements of the form $(x,y)$.

So $(x_1,y_1)$ and $(x_2,y_2)$ are two elements in the set.

You'll have to prove that $(x_3,y_3) = ((1-lambda)x_1+lambda x_2,(1-lambda)y_1+lambda y_2)$ also is an element of the set if $(x_1,y_1)$ and $(x_2,y_2)$ are elements of the set and $0<lambda<1$.

$B = { ... }$

This set could also be written as:

$B={(x,y)inmathbb R:x>0,y>0,(x,y)in A}$

(By the way: Is the notation "$mathbb R$" correct here or should it be "$mathbb R^2$"?)

In the first step you will already have proven that $(x_3,y_3)in A$ if $(x_1,y_1)in A land (x_2,y_2) in A$.

This means you only need to prove that $x_3>0$ and $y_3>0$ if $(x_1,y_1)in B land (x_2,y_2)in B$.

EDIT

It is not obvious what shall to do next

Unfortunately I don't know if I have the right answer here, either. I was thinking that your only problem is understanding the definition of the set.

However I would try to do the following:

The points $(0,1)$ and $(1,0)$ are elements of the set $A$. If the set is convex, the point $(frac{1}{2},frac{1}{2})$ is also an element of the set.

This means:

$(frac{1}{2})^p+(frac{1}{2})^pleq 1\
Rightarrow 2(frac{1}{2})^pleq 1\
Rightarrow 2leq 2^p$

This condition is obviously not true for $p<1$ so the set $A$ can only be convex for $pgeq 1$.

Proving this for the set $B$ will be harder because $(0,1)notin B$...

You may argue with the points $(1-2epsilon,2delta)$, $(2delta,1-2epsilon)$ and with limit calculation: If $delta$ and $epsilon$ are small enough...

To prove that $A$ is convex for all $pgeq 1$ I'd first define a modified variant of the set $B$:

$C={(x,y) : xgeq 0, ygeq 0, |x|^p+|y|^pleq 1}={(x,y) : xgeq 0, ygeq 0, x^p+y^pleq 1}$

The following inequations are true for every element in the set:

$x_1^p + y_1^p leq 1\
x_2^p + y_2^p leq 1$

And because $0leqlambdaleq 1$:

$lambda x_1^p + lambda y_1^p leq lambda\
(1-lambda)x_2^p + (1-lambda)y_2^p leq 1-lambda$

$Rightarrow lambda x_1^p + (1-lambda)x_2^p + lambda y_1^p + (1-lambda)y_2^p leq 1$

We can now have a look at the following function:

$f(x)=x^p, xin[0,infty), p>1\
f'(x)=px^{p-1}geq 0\
f''(x)=p(p-1)x^{p-2}geq 0$

When using the properties of convex and concave functions we see that:

$lambda f(x_1)+(1-lambda) f(x_2)geq f(lambda x_1+(1-lambda)x_2)$

(I don't know if you are allowed to use this in some university homework!)

For $f(x)=x$, which means $p=1$, this is also true.

This means:

$lambda x_1^p+(1-lambda) x_2^pgeq (lambda x_1+(1-lambda)x_2)^p$

Or:

$x_3^p+y_3^p = (lambda x_1 + (1-lambda)x_2)^p + (lambda y_1 + (1-lambda)y_2)^p \ leq lambda x_1^p + (1-lambda)x_2^p + lambda y_1^p + (1-lambda)y_2^p leq 1$

Therefore $(x_3,y_3)in C$.

Now we can also define another set:

$D={(x,y) : ygeq 0, |x|^p+|y|^pleq 1}={(x,y) : (x,y) in C lor (-x,y) in C}$

It should be easy to prove that this set is also convex; we have to distinguish three cases:

• $x_1,x_2geq 0 Rightarrow (x_3,y_3)in C Rightarrow (x_3,y_3)in D$


• $x_1,x_2leq 0 Rightarrow (-x_3,y_3)in C Rightarrow (x_3,y_3)in D$


• 
  $x_1$ and $x_2$ have different signs:
  
  This case is a bit more difficult; we first have to prove that the set contains an element $(0,y_0)$ "between" the points $(x_1,y_1)$ and $(x_2,y_2)$...

Now the set $A$ can be written as: $A={(x,y):(x,y)in Dlor (x,-y)in D}$.

The steps needed to prove that $A$ is convex when knowing that $D$ is convex are the same steps needed to prove that $D$ is convex when knowing that $C$ is convex.

EDIT 2

I just thought about the cases $pleq 0$:

My proof that $A$ is not convex unfortunately will not work for $pleq 0$; however you can easily show that $A$ cannot be convex for $p<0$ because $(x,y)in A Rightarrow (-x,-y)in A$ but $(0,0)notin A$.

However $B$ seems to be convex for $p<0$!